mb instead look at this

There's a great answer B)

@Slereah not enough sources or derivations

-1

:,(

Source was mostly journals.aps.org.sci-hub.club/prd/abstract/10.1103/…

it's gonna stay at -1 until you put that in

fine

you monster

It is added

what about the Hehl-Datta equation

It's the one published by Hehl and Datta

source?

@0celo7 Hehl and Datta, obviously :P

scitation.aip.org/content/aip/journal/jmp/12/7/10.1063/…

put it in the post!

I did :V

where

The link is on the "hehl datta"

nope

now it updated

so what are kepler's laws again?

$T^2/a^3=\mathrm{const.}$

(If there's no torsion it is called the Heisenberg-Ivanenko equation, btw)

equal area in equal time

There's 3 Kepler's laws

yah what's the third one

elliptical orbits?

yes

Kepler basically had all that shit before everyone else

But Galileo didn't care because he was a crazy man

like a 15th century ACM

putting blades in gullets since 1576

so 16th century

The Gordon decomposition is an easy thing to do because it is just replacing $\psi$ by $\frac{-i}{m}gamma \partial \psi$

and bam you have the gordon decomposition

woot

got the overrides

math major official

@0celo7 Not sure if flattered or offended.

why would you be offended?

also I'm going to see the research reactor at ORNL! Cool!

and Denzler is my faculty advisor, nice

what is his research, anyway

> Partial Differential Equations (in particular spectral, geometric, and dynamical systems questions).

he had an awful lot of topology books on his shelves for a PDE guy

everything is topology

lol

fun office

physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf

^pretty good thing on the gordon decomposition btw

I want to say it was an exercise in Zee to derive that

@yuggib

You can see that the spin current is similar in form to the EM tensor

So you can do

$*dF = j_c + j_s$ zoop $*dF - j_s = j_c$

And integrate it into the EM tensor

And that is the POLARIZATION and the MAGNETIZATION

$*d(*F - \bar \psi \sigma \psi) = j_c$

Hm wait

Isn't that a 2-form

If $\bar \psi \sigma \psi$ is a 2-form mb it wouldn't care bout torsion at all

And only the orbital current would be affected by torsion

@Slereah: Use $\star$ not $*$, you savage!

asterisk is used for convolution theorem what else?

complex conjugate

pullback

footnotes

smileys

Also the Hodge star

And the Green star

and whatever star

SAVAGE ALERT

@Slereah SAVAGE

@ACuriousMind sometimes I feel like we're the only civilized ones

::strokes long hippie hair:: Sometimes I think we are.

nipple

nevermind, I'm the only civilized one

@ChrisWhite pls write the Maxwell equations

jesus

savage

$\mathrm{d}^\dagger$ or $\delta$ are acceptable

wha

wtf

what does that mean

@ChrisWhite That's no good because one sometimes defines $\mathrm{d}^*$ to be the adjoint operator of $\mathrm{d}$.

@ChrisWhite Really? :O

Who designed that? :P

@ACuriousMind that's what he meant I think

$\star\mathrm{d}\star$ is the adjoint of $\mathrm{d}$

modulo a shitty sign

Hmmmm

oh is that why $\star\mathrm{d}\star$ looks funny

I retract the objection about the adjoint, but I'm now not exactly sure what the sign in that equation has to be :D

CW's LaTeX skills surpasseth all human understanding

Ah, I know why it's binary: It's supposed to be the Moyal star.

@ACuriousMind just redefine $J$ :)

Moyal star is neat

It's another sacrifice on the pile of quantum formalisms

I am currently making a list of all QFT formalisms

There is a lot

cluster decomp for days

let it be known that ACM was corrected by an engineer

Schrodinger, Heisenberg, quantum action, path integral (Euclidian and other types), phase space quantization, haag QFT, the other axiomatic QFT, Bohm QFT, stochastic QFT

So many QFTs

Bajoran QFT

Cardassian QFT

en.memory-alpha.wikia.com/wiki/Bessel_function

What's that one

Btw Bajoran QFT is the one where a ginger German ninja shanks you if you say quantum fluctuation

I'm not ginger.

But @0celo7

Never said you were

Why would you think that

What is $\langle \hat A \rangle^2 - \langle \hat A^2 \rangle$ then

@0celo7 Because you have repeatedly implied that you think I shank people.

So?

What, you're the only German who can shank people?

@0celo7 Of course, we are peaceful people.

Just him and the Stuttgart murderer

@ACuriousMind really

aight, registered for classes

Quick question, by Faraday's and Ampere's law we know that a spatially varying electric field will induce a time dependent magnetic field too. I have read just before that the wave equations of an electromagnetic wave for the B and E fields are independent of one another. I am struggling to see how the author of the article concludes that the field propagations are independent (ill link the article) .... Although I do see how both B and E fields have their own wave equation

physics.info/em-waves

scroll to .... "Let's clean it up a bit and see what we get."

@ACuriousMind

@0celo7 99% sure they can't. Not 100% sure though, hence no answer.

why can't they

well, I know of no mechanism that would allow them to tangle

but maybe string field theory can produce tangled strings or something

can you make handcuffs with strings

mb we can arrest God

why did you post that picture that is starred

I can't stop laughing at it

it is history

@0celo7 The spatial slices of the worldsheets have to be the strings, essentially. There's no orientable 2D surface where a slice is interlocked strings. But my imagination is not good enough to tell whether the configuration occurs as a slice of a non-orientable surface.

that dude just does. not. give. a. fr*ck.

@ACuriousMind how do you know there is none

orientable that is

@0celo7 All the orientable ones are embeddable into $\mathbb{R}^3$ without intersection. If you think about how the interlocked strings embedded in 3D should extend to a worldsheet, I find it pretty obvious that the resulting worldsheet intersects itself in 3D.

Hm

Is that a theorem that all orientable surfaces can be embedded in R^3?

bah

ACM is just a denier

he doesn't have faith in Lord Witten

I bet he thinks quantum mechanics is correct

@Slereah They're all the connected sum of tori, since the genus invariant classifies them completely, and the torus is embedded in $\mathbb{R}^3$.

damn, I knew that

doesn't mean you're right though

Oh right

Wait

Isn't the classification of 2-manifolds by genus only valid for compact manifolds

do I smell a #rekt

or can ACM recover

@Slereah Closed string worldsheets are compact. The non-compact ones come from open strings, and open strings can't be interlocked, so we should look at compact ones.

Aight

are they compact

why can't you have an infinite cylinder

I have one ( ͡° ͜ʖ ͡°)

@0celo7 Because it's conformally equivalent to a twice-punctured sphere.

just testing you

no really, I did know that

after a bit :)

The terminology is a bit imprecise there because the twice-punctured sphere is not compact (you can't have a diffeo that maps something non-compact to something compact), but it's inside the compact sphere, which is good enough.

I'm never sure which of these we mean exactly when we talk about the "worldsheets", but I suspect the people writing about them aren't either :P

Aaaah

Finally jobless

Worldsheet = pants

RPfnoR 3

@Secret you know string theory?

nothing beyond pop sci, thus you can say no

I do have "A First Course in String Theory" but I have not had the time to read it yet

I wonder if I can get a scholarship from the math department now

you doing a maths subject?

@Secret is that a ghost

especially the left one

(Joke taken too far...) Now see if it also has negative kinetic term and we are in business

That's a phantom not a ghost >:|

ffs

learn some physics kid

@Slereah jobless = time to play gaems

I am still far, but not very far now (compared to 3 years ago) before I will finally be able to meet ghost and phantoms in QFT exercises

phantoms don't show up often

I've never seen one

Phantoms are barely a thing in cosmology

They were mildly popular in the late 90's during the accelerated expansion discovery

your mom was mildly popular in the late 90s :)

Yes she was Avril Lavigne

oooooo

fite her

lol you can select birth year as 2016 when signing up for ORNL HFIR training

"One of the many consequences of the theorem is that any two symplectic manifolds of the same dimension are locally symplectomorphic to one another"

I refuse to believe that sympletomorphic is a word

I do to

it should be symplectomorphic

sympletomorphic is just silly

that should be *too

mom, since ACM doesn't know what worldsheets are and since worldsheets are pants, he doesn't know what pants are?

@ACuriousMind savage

you can \mathrm and \star and define the exerior algebra by quotienting an by an ideal all you want, but pants?

you're a savage in disguise!

fake

so what

@ChrisWhite Is the way that Arnold defines integration of forms "legit"? i.e. is it always possible to find an $f$ that maps a polyhedron in $\mathbb{R}^k$ to the desired chain in $M$?

@ChrisWhite On page 186, Arnold says in order to prove $\partial^2 \sigma=0$ for a $k$-chain $\sigma$, one only has to look at $\partial^2 D$ for a convex polyhedron $D$. Then he says it is enough to prove this ($\partial^2D=0$) for $k=2$. Why?

$\partial$ is the boundary operator

@yuggib you are free to answer as well

@obe you too

funny.

@0celo7 I don't know much about differential forms and stuff like that (assuming we are talking about that o.O)

I mean sad.

you said you read this book

@yuggib second question is more pressing

it's a topology question

@0celo7 I read the first chapter and skimmed through some other parts.

seriously...

whatever fml.

I don't get the $k=2$ piece

;(

@0celo7 A $k$-chain is not a usual topological term

and also "a boundary operator"

they seem quite geometrical to me

if they are topological, simply I do not know them :P

well you're a mathematician

so certainly you can help with an exercise?

I mean it's obvious for $k=2$

@ACuriousMind I need you D:

you just just bump up the dimensions of everything? like points are now $k-2$-dimensional?

lines $k-1$?

etc.?

@0celo7 I am a physicist :-P

and my Ph.D. was in analysis, so really geometry/algebra is not familiar to me...

dude it's like a baby problem

then you should be able to do it easily :P

I'm bad at math :/

Wow why does he define the exterior derivative so complicated?

What's wrong with "the exterior derivative is the unique degree 1 antiderivation on the Grassmann bundle whose action on a function is the differential."

Man I need to sleep

Stop saying weird words

Weird?

I thought you bought Straumann. That's how he defines it.

@ACuriousMind have you taken analysis 1?

@NeuroFuzzy I'm confused by Arnold's construction of the symplectic structure on page 202

@0celo7 I didn't get that far I don't believe

damn

Yeah, nope, that was the point where I realized that despite Arnold's valiant efforts to explain differential forms and stokes and homology, I didn't understand any of it.

or, the chapter before that was*

I think his approach makes it hard

from what I understand, this is the Russian way of doing math

I wish he explained what $p(f_*\xi)$ means

@ACuriousMind !

please!

explain the proof on the bottom of 202!

@ACuriousMind Should that be $\mathbf{p}(f_*\xi)$, viewing the momentum as a one-form acting on the tangent space?

and somehow $\mathrm{d}\mathbf{q}(\xi)=f_*\xi$?

dude I'm confused :/

I will be available shortly. Do not despair.

TOO LATE

is cold, dark and alone

This was Arnold's plan all along...

@gonenc Yes. I very much recommend at least trying it.

@yuggib A $k$-chain is very much a topological term. Homology is defined as the quotient of the $k$-cycles by the $k$-boundaries, both of which are special cases of chains.

::waits for wise words of clarity to be formulated::

@0celo7 I think he says "it is enough" not in the sense of "it is equivalent" but in the sense that the reasoning you apply in 2D is the same you apply in every other dimension.

@ACuriousMind that's what I thought

now on to the other thing

@0celo7 Which one?

page 202!!!!

@0celo7 $p$ is a 1-form on $V$. $f_{*}$ is the derivative/Jacobian (also variously denoted $Df$ or $\mathrm{d}f$) mapping a tangent vector of $T^*V$ to a tangent vector of $V$. Therefore, $f_* \xi$ is a tangent vector to $V$ and $p(f_*\xi)$ is a number.

oh lord

i know all of that

that's not my problem

$\omega^1$ is the tautological one-form

what is $p(.)$ explicitly

and why does $\omega^1$ equal what it does

@0celo7 $\omega^1$ is defined by that equation.

then what the hell is $p$??

wait a moment

what is $m$ in the wiki article

that looks like $p$ here

@0celo7 For any point $x\in T^* V$, if you have coordinates $q$ for $V$, the vector $p$ is given by the action of the 1-form associated to $x$ at $q$.

@0celo7 Exactly.

@ACuriousMind well it doesn't make sense me to me either

how about this

how does he get the form of $\omega^1$ from all of this

@0celo7 $f$ from Arnold is $\pi$ in the Wiki article.

@ACuriousMind yes!

I can read!

Okay :)

but I still don't see how $\omega^1$ comes out at the end

Have some patience with me, I'm not exactly sober ;)

either from Wiki or Arnold

@0celo7 The $p(f_{*}\xi)$ or the $p\mathrm{d}q$?

@ACuriousMind latter

I know the former is the definition

at least it seems so

ok, so a point in the cotangent bundle is just a one-form associated with a point

so by $p(.)$ we mean the one-form at point $p\in T^*V$ acting on a vector at point $q$ on the manifold

@0celo7 Okay. $f_{*}\xi $ is a tangent vector to $V$ and $p$ is a one-form on $V$. So, we can certainly expand $p = p_i \mathrm{d}q^i$. Now, we can expand $\xi = \xi^a \frac{\partial}{\partial q^a} + \xi^b \frac{\partial}{\partial p^b}$. The projection $f_{*}$ kills the latter part, and what remains is using $\mathrm{d}q^i(\frac{\partial}{\partial q^j}) = \delta^i_j$ to arrive at the expression for $\omega^1$.

wow, you are drunk

sigh

once again

had that exact thing on a paper

@0celo7 One has to escape the asterisks properly else the chat thinks they're meant to make stuff italic.

did not believe it

test $f_{} \mu\nu\rho^{}$

indeed!

damn, drunk you is cleverer than sober me

well, thanks for the help

if only to confirm what I had written an hour ago

@0celo7 I think confidence in such things comes really only from doing it over and over again.

@0celo7 :)

ok, so here's where my confusion is

In unrelated news, the lecture "Lorentzian manifolds" is not as nice as I hoped it to be :/

$\mathrm{d}q^i$ is a one form on $V$ and on $T^*V$?

and $p_i$ is a one-form on $V$ and $\mathrm{d}p_i$ one-form on $T^*V$?

@0celo7 Since $q$ are (full) coordinates for $V$ and (part of the) coordinates for $T^*V$, yes.

ok, so regarding the remark

how do we know that $p_i$ is a one-form on $V$?

Arnold says it follows from the definition via the Lagrangian

@0celo7 Not exactly. The $p$ is part of the full coordinate chart $(q,p)$ for $T^*V$, and since the $q$ are full coordinates for the $V$ part of the local $T^*V \cong V\times \mathbb{R}^n$, the $p$ lives, locally, purely in the cotangent space, and hecne is a 1-form on $V$, so one can expand it as $p = p_i\mathrm{d}q^i$.

There is some further thought involved to see that the definition of $p$ indeed makes its components $p_i$ exactly the coefficents of that one-form.

But I'm a bit too lazy now to chase the definition of $p$ back to that - you might as well take the relation $p = p_i \mathrm{d}q^i$ as the defining characteristic of the coordinate chart $p$.

When will you be not too lazy :)

The fact that all of this is reliant on coordinate charts leads one to consider any suitable $\omega^2$ as defining the symplectic structure, by the way, so one has a global definition of that structure that's manifestly coordinate invariant.

@0celo7 Perhaps tomorrow, but no promises.

@ACuriousMind does Arnold give that?

it seems like Arnold is less...just different than other books I've read

@0celo7 Dunno, I just read the pages you ask me to ;)

so the definition here depends on coordinates?

well, it seems the coordinate independent one is stuck away in there

@0celo7 No, it doesn't, but it is not obvious that it doesn't (at least to me).

isn't the equation $\omega^1(\xi)=p(f_*\xi)$ coordinate independent?

@0celo7 It should be, but $p$ is a (partial) coordinate chart for $T^*V$. What's missing is defining the 1-form $p$ associated to $x\in T^* V$ in an independent way.